-
Anonymous users2024-02-11Solution: Ellipse: t 2-1>0, t 2>0
t>1 or t<-1
Hyperbola: t 2>0, t 2-1<0
1>t>-1
Ellipse: Because t 2 > t 2-1
So the focus is on the x-axis.
a^2=t^2,b^2=t^2-1
c^2=a^2-b^2=1
c=1 hyperbola:
Because t 2>>0>t 2-1
So the focus is on the x-axis.
a^2=t^2,b^2=1-t^2
c^2=a^2+b^2=1
c=1 all have the same focus.
-
Anonymous users2024-02-101.Elliptical. t >0, t -1>0 , t >1, -10, t -1<0 , t <1, t <-1 or t>1
The hyperbolic equation is:
x²/t²)-y²/(1-t²)]=1
2.Ellipse: C 2 = A 2-B 2 = T -(t -1) = 1, C = 1, Hyperbola: C 2 = A 2 + B 2 = T + (1-t ) = 1, C = 1
Have the same focus.
-
Anonymous users2024-02-09An ellipse is an x-square and y-square, and the numbers below are not equal and both are greater than 0
Hyperbola is one greater than 0 and the other less than 0
It was too cumbersome, but it wasn't a lot of trouble.
-
Anonymous users2024-02-08Question: 1) The right focus f2 is (1,0), and the middle faction is split.
The half-focal length c is 1, a 2 = b 2 + c 2 = b 2 + 1;
and combine the points m(2,2 10 3) on the ellipse, and substitute the coordinates of a 2 = b 2 + 1 and m into the equation to obtain:
9b^4-67b^2-40=0
Factorization: (9b 2+5)(b 2-8)=0b 2>0, b 2=8, a 2 = 9
The equation is: x 2 9 + y 2 Luga 8=1
2) There is no solution to this problem, and I am still thinking about it.
-
Anonymous users2024-02-07x^2/16-y^2/9=1
A 2 = 16, b 2 = 9, c 2 = 16 + 9 = 25, so the focal coordinates are (-5,0) and (5,0).
i.e. a 2 = b 2 + 25 with an ellipse
Let the elliptic equation be x 2 a 2 + y 2 (a 2-25) = 1p (6, root number 7) and substitute to get 36 a 2 + 7 (a 2-25) = 136 (a 2-25) + 7a 2 = a 4-25a 2a 4-68a 2 + 36 * 25 = 0
a^2-50)(a^2-18)=0
a^2=50,a^2=18
b 2 = 25, b 2 = -7<0 (round).
So the elliptic equation is x 2 50 + y 2 25 = 1
-
Anonymous users2024-02-06Focus (5,0) (-5,0).
The sum of the distances from p to f1 and f2 is 2a=2 2+8 2, a=5 2, so b=5, b2=25, a2=50
x^2/50+y^2/25=1
Hope it helps!
If you don't understand something, please ask!
-
Anonymous users2024-02-05The focal coordinates are (0, root number 3), i.e. there is c = root number 3, and the distance from a focal point to the nearest vertex is root number 3-1, i.e. there is c-a = root number 3-a = root number 3-1
Then there is a=1, b 2=c 2-a 2=3-1=2, so the equation is y 2 1-x 2 2=1
-
Anonymous users2024-02-04Focus coordinates (c,0), vertex coordinates (a,0), so the root number 3-1 is a-c
The upstairs idea is correct, but in the end there is a problem when describing the center of the circle, because the equation is (x+2a) +y = (2m), that is, the center of the circle is on the negative half axis, and point b is on the positive half axis, how can b be the center of the circle, the center of the circle should be at a point with a distance of 2a to the left of a, the others are correct, in fact, the equation is not so complicated, it can be obtained according to the observation of geometric properties, and c is cf ad, because d is the midpoint of bc, then ad=2m, a is (-2a, 0), No matter how C changes, the above conclusion can be obtained by passing C as CF AD, then the point c is on a circle with the center of the circle (-2a,0) and the radius of 2m, so the trajectory equation is (x+2a) +y =(2m).
The straight line l:y = root number 3 (x-2) and the hyperbolic x2 a2-y2 b2 = 1 intersect at two points ab, ab = root number 3, there is a line l about l'y=b ax symmetry line l2 is parallel to the x-axis. >>>More
Method 1: Derivative.
y'=2x+2 >>>More
I feel that this is very difficult to look at, but it is actually easy to walk, when the coach taught me, he didn't say anything to me, let me go by feeling, only say that the direction can not go back to death Generally half a circle, a circle, S bend has two curves, one to the left and one to the right, in fact, we go best by feeling, don't remember what point or the like (personal experience) Turn left you will play the direction to the left, when entering the S bend, let the car lean slightly to the right because the first turn is to turn left, the speed should be slow, so that there is enough time to adjust the direction, After entering the S bend, first hit half a circle to the left and then play half a circle or directly hit a circle direction, sometimes you may need to play more than a circle, then you can stretch your head slightly out of the window to see the distance between the side line and the body, adjust the steering wheel appropriately, the left side is narrow to the right, and vice versa (the left side is wide and the right side is narrow) Note that it is fine-tuned generally a small half circle, after the first turn to the direction of the right, go forward seconds or so to the right to play a circle (or more than a circle) direction, according to the method of the first bend is OK, I took the subject two last month, Feel that the S bend is the easiest, by feeling, practice more, don't remember the point, just look at the distance between the body and the sideline to adjust the steering wheel, as long as the wheel is not pressed the line is qualified, come on!
According to the title, s1=490m, v1=240m s, v2=25m s, g=because the motion is a flat throwing motion, the partial motion of the motion in the horizontal direction is a uniform linear motion, and the partial motion in the vertical direction is a free fall motion; >>>More